Simplify and expand the following expression: $ \dfrac{2}{2a + 20}- \dfrac{2}{4a - 40}- \dfrac{a}{a^2 - 100} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{2}{2a + 20} = \dfrac{2}{2(a + 10)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{2}{4a - 40} = \dfrac{2}{4(a - 10)}$ We can factor the quadratic in the third term: $ \dfrac{a}{a^2 - 100} = \dfrac{a}{(a + 10)(a - 10)}$ Now we have: $ \dfrac{2}{2(a + 10)}- \dfrac{2}{4(a - 10)}- \dfrac{a}{(a + 10)(a - 10)} $ The least common multiple of the denominators is: $ 8(a + 10)(a - 10)$ In order to get the first term over $8(a + 10)(a - 10)$ , multiply by $\dfrac{4(a - 10)}{4(a - 10)}$ $ \dfrac{2}{2(a + 10)} \times \dfrac{4(a - 10)}{4(a - 10)} = \dfrac{8(a - 10)}{8(a + 10)(a - 10)} $ In order to get the second term over $8(a + 10)(a - 10)$ , multiply by $\dfrac{2(a + 10)}{2(a + 10)}$ $ \dfrac{2}{4(a - 10)} \times \dfrac{2(a + 10)}{2(a + 10)} = \dfrac{4(a + 10)}{8(a + 10)(a - 10)} $ In order to get the third term over $8(a + 10)(a - 10)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{a}{(a + 10)(a - 10)} \times \dfrac{8}{8} = \dfrac{8a}{8(a + 10)(a - 10)} $ Now we have: $ \dfrac{8(a - 10)}{8(a + 10)(a - 10)} - \dfrac{4(a + 10)}{8(a + 10)(a - 10)} - \dfrac{8a}{8(a + 10)(a - 10)} $ $ = \dfrac{ 8(a - 10) - 4(a + 10) - 8a} {8(a + 10)(a - 10)} $ Expand: $ = \dfrac{8a - 80 - 4a - 40 - 8a}{8a^2 - 800} $ $ = \dfrac{-4a - 120}{8a^2 - 800}$ Simplify: $ = \dfrac{-a - 30}{2a^2 - 200}$